Question

two electrons are released towards each other with equal velocities of 10 to the power 6 meter second-1. what will be the closest approach between them​

Answer 1

Answer:

2.53 Å

Explanation:

Total kinetic energy of electrons

K.E

=

2

×

1

2

mv

2

Potential energy of electrons

U

=

ke

2

r

When they approached closest distance (

r

) their total kinetic energy is converted into potential energy.

K.E = U

2

×

1

2

mv

2

=

ke

2

r

r

=

ke

2

mv

2

r

=

9

×

10

9

Nm

2

/

C

2

×

(

1.6

×

10

−

19

C

)

2

9.1

×

10

−

31

kg

×

(

10

6

m/s

)

2

=

2.53

×

10

−

10

m

=

2.53 Å